## Precalculus (6th Edition)

$\color{blue}{35(5t+3)^{-5/3}(-5t-2)(3t+2))}$
The least exponent of the binomials is $-5/3$. The numerical coefficients have a greatest factor of $7$. Factor out $7(5t+3)^{-5/3}$ to obtain: $=7(5t+3)^{-5/3}[1+2(5t+3)-3(5t+3)^{2}]$ Distribute $2$ to obtain: $\\=7(5t+3)^{-5/3}[1+10t+6-3(5t+3)^{2}] \\=7(5t+3)^{-5/3}[10t+7-3(5t+3)^{2}]$ Square the binomial to obtain: $\\=7(5t+3)^{-5/3}[10t+7-3(25t^2+30t+9)]$ Distribute $3$ then combine like terms to obtain: $=7(5t+3)^{-5/3}(10t+7-75t^2-90t-27) \\=7(5t+3)^{-5/3}(-75t^2-80t-20)$ Factor out $5$ in the trinomial to obtain: $\\=7(5t+3)^{-5/3}\cdot 5(-15t^2-16t-4) \\=35(5t+3)^{-5/3}(-15t^2-16t-4)$ Factor the trinomial to obtain: $=\color{blue}{35(5t+3)^{-5/3}(-5t-2)(3t+2))}$